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galina1969 [7]
3 years ago
7

Consider a 1.6 × 10-3 M solution of HNO3. Which of the following statements is NOT true? Consider a 1.6 × 10-3 M solution of HNO

3. Which of the following statements is NOT true? This solution could dissolve metal. This solution could neutralize a base. This solution would turn litmus to red. This solution has a pH of 11.20. none of the above
Chemistry
1 answer:
Serjik [45]3 years ago
3 0

Answer:

This solution has a pH of 11.20

Explanation:

HNO3 is a strong acid known to ionize completely in solution. Being a strong acid, a low pH value is expected. As expected, an acid will display the following properties:

This solution could dissolve metal.

This solution could neutralize a base.

This solution would turn litmus to red.

The pH of this solution is obtained from -log[H+]

pH = -log(1.6 × 10-3)= 2.796

Hence the statement; This solution has a pH of 11.20 is not correct.

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Which class of elements will contain both solids and gases at room temperature?
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alculate the standard enthalpy of formation of ethanoic acid given that the standard enthalpy of combustion for carbon is –394 k
agasfer [191]

Answer:

The standard enthalpy of formation of ethanoic acid is -484 kJ/mol.

Explanation:

C(g)+O_2(g)\rightarrow CO_2(g),\Delta H_{1, comb}=-394 kJ/mol...[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l),\Delta H_{2, comb}=-286 kJ/mol..[2]

CH_3COOH(l)+2O_2(g)\rightarrow 2CO_2(g)+2H_2O(l),\Delta H_{3, comb}=-876 kJ/mol..[3]

The standard enthalpy of formation of ethanoic acid :

2C(g)+2H_2(g)+O_2(g)\rightarrow CH_3COOH, \Delta H_{4}=?..[4]

Using Hess's law to calculate :

2 × [1] + 2 × [2] - [3] = [4]

\Delta H_4=2\times (-394 kJ/mol)+2\times (-286 kJ/mol) - (-876 kJ/mol)

=\Delta H_4=-484 kJ/mol

The standard enthalpy of formation of ethanoic acid is -484 kJ/mol.

5 0
3 years ago
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6 0
3 years ago
A large balloon contains 233 L of neon at 22.0C and 760. Torr. If the balloon ascends to an altitude where the temperature is -
shepuryov [24]

Answer: The volume in the balloon at the higher altitude is 260 L

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 760 torr

P_2 = final pressure of gas = 511 torr

V_1 = initial volume of gas = 233 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 22.0^0C=(22.0+273)K=295K

T_2 = final temperature of gas = -52.0^0C=(-52.0+273)K=221K

Now put all the given values in the above equation, we get:

\frac{760\times 233}{295}=\frac{511\times V_2}{221}

V_2=260L

The volume in the balloon at the higher altitude is 260 L

3 0
3 years ago
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